Higgs Mechanism

When a local rather than global symmetry is spontaneously broken, we do not get a massless Goldstone boson. Instead, the gauge ﬁeld of the broken symmetry becomes massive, and the would-be Goldstone scalar becomes the longitudinal mode of the massive vector. This is the Higgs mechanism, and it works for both abelian and non-abelian local symmetries. In the non-abelian case, for each spontaneously broken generator T a of the local symmetry the a corresponding gauge ﬁeld Aµ(x) becomes massive. The Abelian Example

To understand how the Higgs mechanism works, let’s start with the abelian example of a local U(1) phase symmetry. The complete model comprises a complex scalar ﬁeld Φ(x) of electric charge q coupled to the EM ﬁeld Aµ(x); the Lagrangian is

1 µν ∗ µ ∗ = FµνF + DµΦ D Φ V (Φ Φ) (1) L − 4 − where

∗ ∗ ∗ DµΦ(x) = ∂µΦ(x) + iqAµ(x)Φ(x),DµΦ (x) = ∂µΦ (x) iqAµ(x)Φ (x), (2) − and λ V (Φ∗Φ) = (Φ∗Φ)2 + m2(Φ∗Φ). (3) 2 Suppose λ > 0 but m2 < 0, so that Φ = 0 is a local maximum of the scalar potential, while the minima form a degenerate circle

v 2m2 Φ = eiθ, v = − , any real θ. (4) √2 × r λ Consequently, the scalar field Φ develops a non-zero vacuum expectation value Φ = 0, which h i 6 spontaneously breaks the U(1) symmetry of the theory. Were that U(1) symmetry global rather than local, is spontaneous breakdown would lead to a massless Goldstone scalar stemming from the phase of the complex field Φ(x). But for the local U(1) symmetry, the phase of Φ(x)— not just the phase of the vacuum expectation value Φ but the x-dependent phase of the h i dynamical Φ(x) field — can be eliminated by a gauge transform, so the physical consequences of the SSB are more complicated.

1 To see how this works, let’s use polar coordinates in the scalar ﬁeld space, thus

1 iΘ(x) Φ(x) = φr(x) e , real φr(x) > 0, real Θ(x). (5) √2 ×

This field redefinition is singular when Φ(x) = 0, so we should never use it for theories with Φ = 0, but it’s OK for spontaneously broken theories where we expect Φ(x) = 0 almost h i 6 everywhere. In terms of the real fields φr(x) and Θ(x), the scalar potential depends only on the radial field φr,

λ 2 V (Φ) = φ2 v2 + const, (6) 8 r − or in terms of the radial ﬁeld shifted by its VEV, φr(x)= v + σ(x),

2 2 2 2 2 φr v = (v + σ) v = 2vσ + σ , (7) − − 2 λ 2 λv λv λ V = 2vσ + σ2 = σ2 + σ3 + σ4. (8) 8 2 × 2 × 8 ×

At the same time, the covariant derivative DµΦ becomes

iΘ 1 iΘ iΘ e DµΦ = ∂µ φre + iqAµ φre = ∂µφr + φr i∂µΘ + φr iqAµ . (9) √2 × √2 × × Inside the big () on the RHS, the ﬁrst term is real while the other two terms are imaginary, hence 2 2 1 DµΦ = ∂µφr + φr i∂µΘ + φr iqAµ | | 2 × × 2 1 2 φr 2 = ( ∂µφr) + ∂µΘ+ qAµ (10) 2 2 × 2 1 2 (v + σ) 2 = (∂µσ) + ∂µΘ+ qAµ . 2 2 × Altogether,

2 1 2 1 µν (v + σ) 2 = (∂µσ) V (σ) FµνF + ∂µΘ+ qAµ . (11) L 2 − − 4 2 × To understand the physical content of this Lagrangian, let’s expand it in powers of the

2 ﬁelds (and their derivatives) and focus on the quadratic part describing the free particles,

2 2 1 2 λv 2 1 µν v 2 = (∂µσ) σ Fµν F + qAµ + ∂µΘ) . (12) Lfree 2 − 2 × − 4 2 × The first two terms here obviously describe a real scalar particle of positive mass2 = λv2. The other two terms — involving the Aµ(x) and the Θ(x) fields — seem to describe a photon and a scalar field, but in fact describe a massive vector field and no scalars!

To see how this works, note the local U(1) symmetry of the theory, which acts as

′ A (x) = Aµ(x) ∂Λ(x), µ − Φ′(x) = Φ(x) exp(iqΛ(x)), × (13) σ′(x) = σ(x), Θ′(x) = Θ(x) + qΛ(x), for an arbitrary x-dependent Λ(x). Physically, such a local symmetry means that one of the 6 field variables at each x — the real and the imaginary parts of the Φ(x), and the 4 components of the Aµ(x) — is redundant, and we may reduce this redundancy by imposing a gauge-fixing µ condition such as the Coulomb gauge A(x) 0 or the Landau gauge ∂µA (x) 0. When we ∇· ≡ ≡ have a charged scalar field with a non-zero VEV, we may also impose a gauge-fixing condition on that scalar field (instead of the vector field Aµ(x)), thus the unitary gauge

Θ(x) = phase Φ(x) = 0 x. (14) ∀ The unitary gauge is badly singular when the complex field Φ(x) fluctuates around zero, so it should never be used for the gauge symmetries which are NOT spontaneously broken. But when the symmetry IS spontaneously broken by Φ = 0 and the points where Φ(x) vanishes h i 6 are few and far between (if they exist at all), the phase Θ(x) is well-defined almost everywhere, and it is easy to gauge it away by setting Λ(x)=( 1/q)Θ(x) = Θ′(x) = 0. − ⇒ In the unitary gauge, the last two terms in the free Lagrangian (12) become simply

2 2 massive 1 µν q v µ = FµνF + AµA , (15) Lvector − 4 2 × the Lagrangian of a massive vector ﬁeld of mass mv = qv. The scalar Θ(x) is gone from this Lagrangian — it was eliminated by the unitary gauge ﬁxing. For the same reason, the

3 Lagrangian (11) is NOT gauge invariant — we used up the gauge symmetry of the original theory for eliminating the Θ(x) ﬁeld, and now the remaining Aµ(x) ﬁeld does not have any gauge symmetry anymore.

Without the unitary gauge — or any other gauge-ﬁxing condition — we may describe exactly the same massive vector particles using redundant ﬁelds Aµ(x) and Θ(x) subject to gauge symmetry

′ ′ A (x) = Aµ(x) ∂µΛ(x), Θ (x) = Θ(x) + qΛ(x), (16) µ − and a gauge-invariant free Lagrangian

2 massive 1 µν (qv) −1 2 = FµνF + Aµ + q ∂µΘ . (17) Lvector − 4 2 × But the Θ(x) ﬁeld here is not physical, it does not give rise to any scalar particles, and its plane waves are mere gauge artefacts. The only physical particles in this system are the massive vector particles, the same as in the Θ-less unitary-gauge Lagrangian (15).

Altogether, the complete particle spectrum of the theory of Φ(x) and Aµ(x) ﬁelds with a spontaneously-broken local U(1) symmetry comprises a massive real scalar σ(x) and a massive vector. But there is NO massless Goldstone scalar!

To see what happened to the would-be Goldstone boson, let’s count the degrees of freedom of the complete theory. The complex scalar field Φ(x) carries 2 degrees of freedom, while the vector field Aµ(x) subject to gauge symmetry carries another 2 DoF, for the total of 4 DoF. This means that for every momentum 3-vector k, there should be 4 distinct 1-particle states k, ?? belonging to different particle species or different spin/polarization states. This counting | i should work for both spontaneously-broken or unbroken U(1) symmetry, although the specific 1-particle states turn out to be quite different for the two regimes:

The unbroken U(1) regime for m2 > 0 and Φ = 0: • h i In this regime, the Aµ(x) ﬁelds describe a massless photon, which has 2 helicity states, λ = 1 (but not λ = 0). At the same time, the complex scalar ﬁeld Φ(x) with an ± unbroken U(1) symmetry describes 2 scalar particle species with opposite electric charges q, the particle and the antiparticle. Altogether, for each k there are 4 one-particle ±

4 states: the scalar particle S+ , the antiparticle S− , and two photon states γ(λ = +1) | i and γ(λ = 1) . | − i The spontaneously-broken U(1) regime for m2 < 0 and Φ = 0: • h i 6 In this regime, there is only one scalar particle species σ, but the massive vector particle has 3 spin states, λ = 1, 0, +1. Again, altogether there are 4 1-particle states: the σ , − | i and the V (λ = +1) , V (λ = 0) , V (λ = 1) . | i | i | − i ⋆ But these are rather diﬀerent 4 states from the unbroken U(1) regime!

Now we can see what happens in the spontaneously-broken regime to the would-be Gold- stone boson Θ(x): It became the longitudinal λ = 0 polarization of the massive vector ﬁeld! Indeed, the unbroken-symmetry regime has a massless vector without the λ = 0 mode. Once the symmetry is spontaneously broken and the vector becomes massive, it has to have all 3 spin states, including the λ = 0 longitudinal mode. That mode has to come from somewhere, so the Higgs mechanism ‘eats up’ the would-be Goldstone scalar Θ(x) and turns it into the longitudinal polarization of the massive vector!

A rigorous way to see how this works would be to start with the redundant gauge-invariant description (16) or a massive vector field, fix the Coulomb gauge A = 0 instead of the unitary ∇· gauge, expand the Lagrangian (17) into Fourier and helicity modes, eliminate the modes of 0 Θ the A field, quantize the theory canonically, and in the process see how the Θˆ k and the Πˆ k operators combine into the creation and annihilation operators for the longitudinally polarized vector particles. But this is a lot of work, and I am not going to do it here. Instead, I let the unitary gauge speak for the outcome of the Higgs mechanism, even if it hides the gory details of the ‘eating up the Goldstone boson’.

To complete this section, let me write down the complete Lagrangian of the spontaneously- broken theory in the unitary gauge, including all the interactions of the σ(x) ﬁelds with itself and with the massive vector ﬁeld:

2 2 2 1 2 λv 2 1 µν q v µ = (∂µσ) σ FµνF + AµA L 2 − 2 × − 4 2 × 2 (18) λv 3 λ 4 2 µ q 2 µ σ σ + qv σ AµA + σ AµA . − 2 × − 8 × × 2 ×

5 Propagators and Gauges for Massive Vectors

The unitary gauge is very useful for identifying the physical particles of the theory. This will be particularly convenient in the Higgsed non-abelian gauge theories, as we shall see later in these notes. But in perturbation theory, the unitary gauge is useful at the tree level but in loop diagrams it makes the UV divergences much worse than in the massless theory. To avoid this problem, perturbation theory beyond the tree level is usually done using non-unitary

Feynman-like Rξ gauges I shall explain in a minute.

To see the problem with loop diagrams, consider the massive photon’s propagator. In the unitary gauge, the massive photon is simply an ordinary massive vector ﬁeld with free Lagrangian (15), so its propagator is just the massive vector propagator from homework#4,

µ i kµkν ν = gµν + for m = qv. (19) k2 m2 + i0 − m2 −

Note that at large oﬀ-shell momenta k m this propagator behaves as O(1/m2), unlike the ≫ massless photon’s propagator which behaves as O(1/k2). So when the propagator (19) is a part of some loop and we integrate over the loop momentum, we get

d4k massive photon other propagators , (20) (2π)4 propagator × and vertices Z the integrand decreases for kµ much slower than for a similar diagram involving a → ∞ massless photon. Consequently, the integral (20) generally suﬀers from much worse ultraviolet divergence — the divergence for k — than a similar integral in a massless QED. In the →∞ QFT (II) class we shall learn how handle the UV divergences in QED and other renormalizable theories. The worse divergences due to the massive photon propagator (19) in the unitary gauge would break the renormalization techniques and make the massive theory non-renormalizable. In other words, it would be a good eﬀective low-energy theory for the tree-level calculations but unsuitable for the loop calculations.

Fortunately, the problem is not with the massive photons per se but rather with the unitary gauge. Other gauges exist where the massive photon propagator behaves like O(1/k2) for large oﬀ-shell momenta, and in those gauges the UV divergences of the massive theory are similar to its massless counterpart.

6 Indeed, consider a non-unitary gauge in which the would-be Goldstone scalar π(x) = µ Im Φ(s) is not restricted at all; instead, we impose a Landau gauge condition ∂µA (x) 0 on ≡ the vector ﬁeld itself. Or rather, impose a Landau-like gauge condition

µ ∂µA (x) ξmπ(x) = given f(x), (21) − and then in the path-integral formulation of the theory (which we shall learn in the QFT II class), we average over all possible f(x). The result of such gauge-averaging is equivalent to not imposing any gauge condition at all but rather adding a gauge-ﬁxing

2 1 µ ∆ = ∂µA ξmπ (22) L −2ξ − to the Lagrangian of the theory. For a massless theory (no Higgs mechanism), the net quadratic Lagrangian

1 µν = FµνF + ∆ (23) L2 − 4 L yields the Feynman gauge (or Feynman-like gauges for ξ = 1) for the photon propagator 6 µ i kµkν ν = gµν + (1 ξ) . (24) k2 + i0 − − k2 + i0 For the massive photon due to Higgs mechanism, we combine the vector ﬁeld Aµ(x) and the would-be Goldstone scalar π(x) with free Lagrangian

1 µν 1 2 2 = 4 Fµν F + 2 (∂µπ) + ∆ L − L (25) 1 2 1 −1 λ 2 1 2 µ 1 2 1 2 2 = (∂µAν) + (1 ξ )(∂ A ) + m AµA + (∂µπ) ξm π . − 2 2 − λ 2 2 − 2

Consequently, we get the so-called Rξ gauge in which the massive vector ﬁeld has propagator

µ ν µν µ ν i µν k k ig = g + (1 ξ) − . k2 m2 + i0 − − k2 ξm2 + i0 −−→ξ=1 k2 m2 + i0 − − − (26) Unlike the unitary gauge propagator (19), this propagator behaves as 1/k2 for large oﬀ-shell momenta k, so the renormalization works similar to QED. (Although the technical aspects of renormalization are much more complicated in the Higgsed case, and it took much longer to prove the theory’s renormalizability.)

7 The price of a non-unitary gauge like Rξ is that the unphysical would-be Goldstone ﬁeld π(x) is not eliminated for the theory. Instead, it remains in the theory, couples to the physical Higgs scalar σ(x), and perhaps to the other ﬁelds (if the theory has any). In Feynman rules, the π(x) has a propagator

i = . (27) k2 ξm2 + i0 −

Note the pole at k2 = ξm2, which is not the physical mass2 of any particle unless ξ = 1; this spurious pole is needed to cancel the other unphysical eﬀects of the massive photon propagator (26).

To summarize this section, loop calculations requires gauges like Rξ where the Feynman rules involve both physical and unphysical fields. However, the tree-level calculation can also be done in the unitary gauge which involves only the physical fields. And the semi-classical calculations where we identify the particles and find their masses — as we shall do in a moment for the non-abelian Higgs mechanism — are best done in the unitary gauge.

Non-Abelian Higgs Mechanism Example: SU(2) with a Higgs Doublet

To illustrate the non-abelian Higgs mechanism, consider the example of SU(2) gauge theory coupled to a doublet of complex scalar ﬁelds Φi(x). In terms of canonically normalized ﬁelds, the Lagrangian is

2 2 1 a aµν ∗ µ i λ ∗ i v = F F + DµΦ D Φ Φ Φ , (28) L − 4 µν i − 2 i − 2 where i i i a a i j DµΦ = ∂µΦ + 2 gAµ σ jΦ , ∗ ∗ i a ∗ a j DµΦ = ∂µΦ gA Φ σ , (29) i i − 2 µ j i a a a abc b c F = ∂µA ∂ν A gǫ A A . µν ν − µ − µ ν For v2 > 0 the scalar potential has a local maximum at Φi = 0 while the minima form a ∗ i 2 2 4 spherical shell Φi Φ = (v /2) in the C = R ﬁeld space; all such minima are related by the

8 SU(2) symmetries to

v 0 Φ = . (30) h i √2 × 1 !

Note that this vacuum expectation value spontaneously breaks the SU(2) symmetry down to nothing — there is no subgroup of SU(2) which leaves this VEV invariant. Consequently, we a expect all 3 vector ﬁelds Aµ(x) to become massive.

In the process, 3 would-be Goldstone scalars should be eaten by the Higgs mechanism. Since the theory has 2 complex — or equivalently 4 real — scalars, only one real scalar should survive un-eaten. Ironically, it is this un-eaten scalar σ(x) which is called the physical Higgs ﬁeld.

To see how this works, let’s ﬁx the unitary gauge. Any complex doublet Φi(x) can be rotated by some SU(2) symmetry U(x) so that the upper component of the rotated Φ′ = UΦ is zero, Φ′1 = 0, while the lower component Φ′2 is real and positive. Thus, in the unitary gauge we require

Re Φ1(x) Im Φ1(x) Im Φ2(x) 0, ≡ ≡ ≡ 1 0 (31) hence Φ(x) = for a real φr(x) > 0. √2 φr(x) !

This gauge-ﬁxing condition is terribly singular for φr 0, so it should never be used for → the unbroken-symmetry regime of the theory. But for the spontaneously broken theory where

φr(x) ﬂuctuates around the minimum at φr = v > 0, the unitary gauge is OK.

In the unitary gauge, the only scalar ﬁeld is the φr(x), or equivalently the shifted ﬁeld

σ(x)= φr(x) v; all the other scalar fields are frozen by the gauge-fixing conditions (31). In − terms of physical Higgs field σ(x), the scalar potential becomes

2 2 2 λ v λ 2 λv λv λ V = Φ†Φ = 2vσ + σ2 = σ2 + σ3 + σ4 (32) 2 − 2 8 2 × 2 × 8 × where the ﬁrst terms is the mass term, mass2 = λv2, while the remaining terms are self-

9 interactions. More interestingly, the covariant derivative of the Higgs doublet Φ becomes

0 ig 3 +1 0 0 + Aµ 1  ∂µσ ! 2 × 0 1 ! v + σ !  DµΦ = − √2  0 1 0 0 i 0   ig 1 ig 2   + Aµ + Aµ −   2 × 1 0 ! v + σ ! 2 × i 0 ! v + σ !      i 1 2 1 g Aµ iAµ (v + σ) = 2 − × , i 3 √2 ∂µσ g Aµ (v + σ) ! − 2 × (33) hence

† µ 1 i 1 2 2 1 i 3 2 DµΦ D Φ = g A iA (v + σ) + ∂µσ g A (v + σ) 2 2 µ − µ × 2 − 2 µ × (34) 2 2 2 2 g (v + σ) 1 2 2 2 g (v + σ) 3 2 1 2 = A + A + A + (∂µσ) . 8 × µ µ 8 × µ 2 The last term here is the kinetic term for the Higgs scalar σ(x), while the rest of the bottom line are mass terms for the vector ﬁelds and the interaction terms between the vectors and a the σ. Curiously, we get the same mass and similar interactions for all 3 vector ﬁelds Aµ:

g2(v + σ)2 M 2 g2v g2 AaAaµ = AaAaµ + σAa Aaµ + σ2Aa Aaµ (35) L ⊃ 8 µ 2 × µ 4 × µ 8 × µ where g2v2 M 2 = . (36) 4

Example: SU(2) with a Higgs Triplet

Now consider an example of a partially broken gauge symmetry, an SU(2) Higgsed down to a U(1) subgroup, or equivalently SO(3) SO(2). This time, the scalar ﬁelds Φa(x) are → real and form a triplet of the SU(2) rather than a doublet. Thus,

1 a aµν 1 a µ a λ a a 2 2 = F F + DµΦ D Φ Φ Φ v , (37) L − 4 µν 2 − 8 − where

a a abc b c a a a abc b c DµΦ = ∂µΦ gǫ A Φ , F = ∂µA ∂νA gǫ A A . (38) − µ µν ν − µ − µ ν Again, for v2 > 0 the scalar potential V (Φ) has a degenerate family of minima which form a

10 spherical shell ΦaΦa = v2 in the scalar ﬁeld space R3, and all such minima are equivalent by

SU(2) ∼= SO(3) symmetries to

0 Φ = 0 . (39) h i    v   

This time, this vacuum expectation value is invariant under an SO(2) subgroup of the SO(3),

— or equivalently under an U(1) subgroup of the SU(2). Specifically, it’s the SO(2) ∼= U(1) generated by the T 3, the third component of the isospin T. Consequently, out of the 3 vector a 3 1,2 fields Aµ, we expect the Aµ to remain massless while the other 2 fields Aµ should become massive.

In the process, the Higgs mechanism should eat 2 real scalar ﬁelds. Since we only have 3 real scalars to begin with, only one scalar should survive un-eaten — the Physical Higgs ﬁeld σ(x).

To see how this works, we ﬁx the unitary gauge

Φ1(x) Φ2(x) 0, Φ3(x) > 0. (40) ≡ ≡

As usual, this gauge is badly singular at Φ = 0, but it’s OK for the Φ(x) Φ = 0. Shifting ≈h i 6 the Φ3(x) by the VEV, we get Φ3(x) = v + σ(x), where σ(x) is the physical Higgs scalar — and also the only scalar remaining in the theory in the unitary gauge.

In terms of the σ(x), the scalar potential becomes

2 λ 2 λv λv λ V (σ) = 2vσ + σ2 = σ2 + σ3 + σ4, (41) 8 2 × 2 × 8 × where the ﬁrst term on the RHS gives the Higgs scalar mass2 = λv2. More interestingly, the

11 covariant derivative of the scalar triple Φa(x) becomes

1 0 Aµ 0 a 2 DµΦ = 0 g A 0   −  µ  ×   3  ∂µσ   Aµ   v + σ        where is the cross product of two isovectors (42) hh × ii gA2 (v + σ) − µ 1 =  +gAµ(v + σ)  ,  ∂µσ    hence the covariant kinetic terms for the scalars become

2 2 1 a µ a 1 2 g (v + σ) 1 2 2 2 DµΦ D Φ = (∂µσ) + A + A . (43) 2 2 2 × µ µ As usual, the ﬁrst term here is the kinetic term for the physical Higgs scalar σ, while the second term contains the mass terms for the vector ﬁelds,

2 M 2 2 A1 + A2 , M 2 = g2v2, (44) L ⊃ 2 × µ µ 1 2 3 but only for the Aµ and the Aµ — the third vector Aµ(x) remains massless.

3 The massless vector Aµ(x) is the gauge field of the un-Higgsed SO(2) ∼= U(1) subgroup of the SO(3) ∼= SU(2). Interpreting this gauge field as the EM field and hence the rescaled generator Q = gT 3 as the electric charge operator, we find that the physical Higgs field is electrically neutral while the massive vector fields have electric charges q = g. To be precise, ± 1 2 the massive vector fields of definite charges are not the Aµ and the Aµ themselves but rather their linear combination

+ 1 1 2 − 1 1 2 Wµ = Aµ iAµ and Wµ = Aµ + iAµ of charges q = g. (45) √2 − √2 ± For completeness sake, let’s re-express the theory at hand (usually called the Georgi–Glashow

12 model) in terms of the physical ﬁelds of deﬁnite charges. Using U(1)–covariant derivatives

± ± 3 ± DµW = ∂µW igA W , (46) ν ν ± µ ν e we have

± def 1 1 2 ± ± Wµν = Fµν iFµν = DµWν DνWµ , (47) √2 ∓ − e e but

3 + − 3 3 F = Fµν + 2g Im W W where Fµν = ∂µA ∂ν A . (48) µν µ ν ν − µ e e Consequently, the Lagrangian of the whole model — the kinetic terms, the mass terms, and the interactions — can be expressed as

1 2 1 2 2 1 µν 1 + −µν 2 + −ν = (∂µσ) M σ FµνF W W + M W W L 2 − 2 σ × − 4 − 2 µν W µ λv λ σ3 σ4 + 2gv σ W +W −µ + g2 σ2 W +W −µ − 2 × − 8 × e ×e × µ × × µ (49) 2 +µ −ν 2 +µ −ν g Fµν Im W W g Im W W . − × × − × e

General Case

Let’s take a closer look at eqs. (34) and (43), and focus on the mass terms for the vector

fields. In both cases, we start with the kinetic terms for the original scalar fields Φi(x) or Φa(x), fix the unitary gauge, work through the algebra, and eventually obtain the kinetic term for the physical Higgs field σ, the mass terms for the vector fields — or some of the vector fields — and the interactions between the massive vectors and the Higgs σ. But if all we want are the mass terms for the vectors, we may simply freeze σ(x) 0: This would eliminate the ≡ 1 2 interactions with the σ as well as the 2 (∂µσ) term, and all we would have left are the mass terms for the massive vectors.

Note that freezing σ(x) 0 is equivalent to freezing all the scalars at their vacuum ex- ≡ pectation values, Φ(x) Φ . Consequently, to get the vector’s masses we do not need to ≡ h i

13 go through the details of the unitary gauge ﬁxing, all we need are the scalar VEVs, then the kinetic terms for the frozen scalars

† 1 2 (Dµ Φ ) (Dµ Φ ) or (Dµ Φ ) h i h i 2 h i become the mass terms for the vectors. For example, for the SO(3) triplet of real scalar ﬁelds from the second example

a abc b c3 ab3 b Dµ Φ = gǫ A vδ = gvǫ A , (50) h i − µ × − × µ

vector 1 a 2 1 2 ab3 ac3 b cµ = Dµ Φ = (gv) ǫ ǫ A A Lmass 2 h i 2 × µ 2 = 1 M = gv A1 A1µ + A2 A2µ . (51) 2 × µ µ Likewise, for the SU(2) doublet of complex scalar ﬁelds from the ﬁrst example,

i ig a a i v j igv a a i Dµ Φ = Aµσ δ = Aµσ , (52) h i 2 j × √2 2 2 (2) × 2 p ∗ igv a a 2 Dµ Φ = Aµσ , (53) h ii −2√2 × i 2 2 vector ∗ µ i g v a a 2 bµ b i = Dµ Φ D Φ = A σ A σ Lmass h ii h i 8 × µ i 2 2 2 g v 2 = Aa Abµ σaσb = δab iǫab3 8 × µ × 2 − g2v2 h i = Aa Abµ δab since Aa Abµ is symmetric in a b. 8 × µ × hh µ ↔ ii M 2 gv = Aa Aaµ for M = . (54) 2 × µ 2

This recipe — freezing Φ(x) Φ to find the vector masses — applies to any kind of gauge ≡h i theory with scalars in any kinds of multiplets. Indeed, consider a general gauge symmetry G â a α with generators T and gauge fields Aµ(x) (a = 1,..., dim(G)). Let scalars Φ (x) belonging to some multiplet (m) of G develop non-zero vacuum expectation values Φα = 0. Then the h i 6

14 covariant derivatives of these scalars

α α a a α β DµΦ (x) = ∂µΦ (x) + igA (x) T Φ (x) (55) µ × (m) β become in the unitary gauge

α α DµΦ (x) = Dµ Φ + terms involving the physical scalars (56) h i where

α a a α β Dµ Φ = igA (x) T Φ . (57) h i µ × (m) β h i In eq. (56), the terms involving the physical scalars — and the physical scalar fields themselves — depend on the details of the unitary gauge fixing. On the other hand, the covariant derivatives of the VEVs (57) depend only on the VEVs themselves. Moreover, such derivatives are linear functions of the vector fields with constant coefficients, so their squares become quadratic mass terms for the vectors,

µ ∗ α a ∗ a β bµ a α γ D Φ Dµ Φ = igA Φ T igA T Φ h iα h i − µ ×h iβ (m) α × × (m) γ h i a bµ 2 ∗ a b β γ = AµA g Φβ T(m)T(m) Φ × h i γ h i (58) by a b symmetry of the Aa Abµ hh ↔ µ ii β = 1 Aa Abµ g2 Φ ∗ T a , T b Φ γ . 2 µ × h iβ (m) (m) γ h i In other words,

ab vector = 1 M 2 Aa Abµ , (59) Lmasses 2 V × µ where the mass2 matrix for the gauge ﬁelds obtains as

ab β M 2 = g2 Φ β T a , T b Φ γ g2 Φ † T a , T b Φ . (60) V h i ∨ (m) (m) γ h i ≡ h i (m) (m) h i

To be precise, eq. (60) applies to Higgs VEVs belonging to a single multiplet of complex 1 scalars. For a multiplet of real scalars, there is an extra factor 2 due to diﬀerent normalization

15 of the VEVS, and for several Higgs multiplets with non-zero VEVs, the general formula is

complex real Higgs Higgs multiplets multiplets ab M 2 = g2 Φ † T a , T b Φ + g2 1 Φ ⊤ T a , T b Φ . (61) V h i (m) (m) h i 2 h i (m) (m) h i Φ∈(m) Φ∈(m) X X

In general, such mass2 matrix is not diagonal, and we need to diagonalize in order to find the physical vector masses. For example, in the Glashow–Weinberg–Salam theory of the weak and EM interactions — it’s explained in the next set of notes — the mass matrix mixes the SU(2) 3 gauge field Wµ and the U(1) gauge field Bµ, and the mass eigenstates are the massless EM

ﬁeld Aµ and the massive neural ﬁeld Zµ involved in the weak interactions.

An additional complication of the GWS theory — or any other theory with non-simple gauge group G = G G — are diﬀerent gauge couplings g for diﬀerent factors G. In 1 × 2 ×··· this case, the g2 factor in eq. (61) for the mass2 matrix element (M 2)ab should be replaced with g(a) g(b) where g(a) is the coupling of the gauge group factor containing the generator T a, × and likewise for the g(b). Thus, the most general formula for the vector mass matrix stemming from the Higgs mechanism is

complex real Higgs Higgs multiplets multiplets  ab 1 M 2 = g(a) g(b) Φ † T a , T b Φ + Φ ⊤ T a , T b Φ . V ×  h i (m) (m) h i 2 h i (m) (m) h i  Φ∈(m) Φ∈(m)   X X         (62) In my notes on the GWS theory we shall see how this works in detail, and how the gauge couplings aﬀect the eigenstates of the mass matrix.